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java - 使用/Java Web 服务器返回文件时出错

转载 作者：行者123 更新时间：2023-12-02 07:58:01 24 4
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我在使用这个用 Java 编写的基本的、不复杂的 Web 服务器时遇到了麻烦。由于某种原因，它们不是仅向浏览器发送“200 OK”或“404 Not Found”，而是写入正在检索的任何文件中。例如，如果我尝试获取 index.html 文件，则会返回:

example

....而不是浏览器尝试编译 HTML。当尝试获取图像时，情况更糟，因为服务器尝试向其附加“200 OK”和内容类型，导致文件损坏。任何人都可以提供有关如何与实际 HTML/JPG 文件分开发送状态行和内容类型的任何见解吗？当我注释掉“os.writeBytes(statusLine)”等时，错误完全消失，但我仍然想将这些消息发送到浏览器......只是不与文件合并。任何帮助将不胜感激。

import java.io.*;
import java.net.*;
import java.util.*;

final class HttpRequest implements Runnable {
    final static String CRLF = "\r\n";
    Socket socket;

    // Constructor
    public HttpRequest(Socket socket) throws Exception {
        this.socket = socket;
    }

    // Implement the run() method of the Runnable interface.
    public void run() {
        try {
            processRequest();
        } catch (Exception e) {
            System.out.println(e);
        }
    }

    private static void sendBytes(FileInputStream fis, OutputStream os) 
    throws Exception {
    // Construct a 1K buffer to hold bytes on their way to the socket.
    byte[] buffer = new byte[1024];
    int bytes = 0;

    // Copy requested file into the socket's output stream.
    while((bytes = fis.read(buffer)) != -1 ) {
        os.write(buffer, 0, bytes);
        }
    }

    private static String contentType(String fileName) {
    if(fileName.endsWith(".htm") || fileName.endsWith(".html")) {
        return "text/html";
    }
    if(fileName.endsWith(".jpeg") || fileName.endsWith(".jpg")) {
    return "image/jpeg";
    }
    if(fileName.endsWith(".gif")) {
    return "image/gif";
    }
    return "application/octet-stream";
    }

    private void processRequest() throws Exception {
        // Get a reference to the socket's input and output streams.
        InputStream is = socket.getInputStream();
        DataOutputStream os = new DataOutputStream(socket.getOutputStream());

        // Set up input stream filters.
        BufferedReader br = new BufferedReader(new InputStreamReader(is));

        // Get the request line of the HTTP request message.
        String requestLine = new String(br.readLine());

        // Display the request line.
        System.out.println();
        System.out.println(requestLine);

        // Get and display the header lines.
        String headerLine = null;
        while ((headerLine = br.readLine()).length() != 0) {
            System.out.println(headerLine);
        }

    // Extract the filename from the request line.
    StringTokenizer tokens = new StringTokenizer(requestLine);
    tokens.nextToken(); // skip over the method, which should be "GET"
    String fileName = tokens.nextToken();
    // Prepend a "." so that file request is within the current directory.
    fileName = "." + fileName;

    // Open the requested file.
    FileInputStream fis = null;
    boolean fileExists = true;
    try {
    fis = new FileInputStream(fileName);
    } catch (FileNotFoundException e) {
    fileExists = false;
    }

    // Construct the response message.
    String statusLine = null;
    String contentTypeLine = null;
    String entityBody = null;
    if (fileExists) {
    statusLine = "200 OK" + CRLF;
    contentTypeLine = "Content-type: " + 
        contentType( fileName ) + CRLF;
    } else {
    statusLine = "404 NOT FOUND" + CRLF;
    contentTypeLine = "Content Not Found!" + CRLF;
    entityBody = "<HTML>" + 
        "<HEAD><TITLE>Not Found</TITLE></HEAD>" +
        "<BODY>Not Found</BODY></HTML>";
    }

    // Send the status line.
    os.writeBytes(statusLine);

    // Send the content type line.
    os.writeBytes(contentTypeLine);

    // Send a blank line to indicate the end of the header lines.
    os.writeBytes(CRLF);

    // Send the entity body.
    if (fileExists) {
    sendBytes(fis, os);
    fis.close();
    } else {
    os.writeBytes("File DNE: Content Not Found!");
    }

        // Close streams and socket.
        os.close();
        br.close();
        socket.close();
    }
    public static void main(String[] args) throws Exception {
        final ServerSocket ss = new ServerSocket(6789);
        while (true)
            new HttpRequest(ss.accept()).run();
    }
}

最佳答案

我相信您缺少一个 CRLF。状态行末尾是否需要额外添加一个

200 好内容类型:xxxx

当然，您可以阅读 HTTP 规范，或者进行wireshark 跟踪:)

更新:是的，你有很多问题。阅读 Easy Http对于一个简单的总结。

但是无论如何你需要说“HTTP/1.0 200 OK”例如

http://www.somehost.com/path/file.html

首先打开一个到主机www.somehost.com的套接字，端口80(使用默认端口80，因为URL中没有指定)。然后，通过套接字发送如下内容:

GET /path/file.html HTTP/1.0
From: someuser@jmarshall.com
User-Agent: HTTPTool/1.0
[blank line here]

服务器应该响应如下内容，通过同一个套接字发回:

HTTP/1.0 200 OK
Date: Fri, 31 Dec 1999 23:59:59 GMT
Content-Type: text/html
Content-Length: 1354

<html>
<body>
<h1>Happy New Millennium!</h1>
(more file contents)
  .
  .
  .
</body>
</html>

关于java - 使用/Java Web 服务器返回文件时出错，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/9372023/

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